What is the integral of (x^3)/((x^2-4)^{3/2)} ?

The integral of (x^3)/((x^2-4)^{3/2)} is (x^2-8)/((x^2-4)^{1/2)}+C

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integral of (x^3)/((x^2-4)^{3/2)}

\int \frac{x ^{3}}{( x ^{2} - 4 ) ^{\frac{3}{2}}} d x

\frac{x ^{2} - 8}{( x ^{2} - 4 ) ^{\frac{1}{2}}} + C

Solution steps

\int \frac{x ^{3}}{( x ^{2} - 4 ) ^{\frac{3}{2}}} d x

Apply u-substitution

: \int \frac{u + 4}{2 u ^{\frac{3}{2}}} d u

= \int \frac{u + 4}{2 u ^{\frac{3}{2}}} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int \frac{u + 4}{u ^{\frac{3}{2}}} d u

Expand

\frac{u + 4}{u ^{\frac{3}{2}}} : \frac{1}{u ^{\frac{1}{2}}} + \frac{4}{u ^{\frac{3}{2}}}

= \frac{1}{2} \cdot \int \frac{1}{u ^{\frac{1}{2}}} + \frac{4}{u ^{\frac{3}{2}}} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{2} (\int \frac{1}{u ^{\frac{1}{2}}} d u + \int \frac{4}{u ^{\frac{3}{2}}} d u)

\int \frac{1}{u ^{\frac{1}{2}}} d u = 2 u^{\frac{1}{2}}

\int \frac{4}{u ^{\frac{3}{2}}} d u = - \frac{8}{u ^{\frac{1}{2}}}

= \frac{1}{2} (2 u^{\frac{1}{2}} - \frac{8}{u ^{\frac{1}{2}}})

Substitute back

u = x^{2} - 4

= \frac{1}{2} (2 (x^{2} - 4)^{\frac{1}{2}} - \frac{8}{( x ^{2} - 4 ) ^{\frac{1}{2}}})

Simplify

\frac{1}{2} (2 (x^{2} - 4)^{\frac{1}{2}} - \frac{8}{( x ^{2} - 4 ) ^{\frac{1}{2}}}) : \frac{x ^{2} - 8}{( x ^{2} - 4 ) ^{\frac{1}{2}}}

= \frac{x ^{2} - 8}{( x ^{2} - 4 ) ^{\frac{1}{2}}}

Add a constant to the solution

= \frac{x ^{2} - 8}{( x ^{2} - 4 ) ^{\frac{1}{2}}} + C

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What is the integral of (x^3)/((x^2-4)^{3/2)} ?
The integral of (x^3)/((x^2-4)^{3/2)} is (x^2-8)/((x^2-4)^{1/2)}+C